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OpenStudy (anonymous):

Find the equation of the tangent to the graph at the indicated point. HINT [Compute the derivative algebraically] f(x) = x^2 − 1; a = 1 I found the slope using lim as h approaches 0, f(1+h) - f(1) / h and got lost during that part.

OpenStudy (anonymous):

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)- f(x)}{h}\] for x = 1: \[ \lim_{h \to 0} \frac{f(1+h)- f(1)}{h} \lim_{h \to 0} \frac{(1+h)^2 -1 - (1^2 - 1)}{h} = \lim_{h \to 0} \frac{1+2h + h^2 -1 - 1 + 1)}{h} = \] \[ \lim_{h \to 0} \frac{2h+h^2}{h} = \lim_{h \to 0} \frac{h(2+h)}{h} = \lim_{h \to 0} \,\,(2+h) = 2 \]

OpenStudy (anonymous):

\[ y-y_1 = m(x - x_1) \] \[ y_1 = f(1) = 1^2 - 1= 0\] equation of the tangent line at that point therefore is \[ y = 2(x-1) \] \[ y = 2x-2 \]

OpenStudy (anonymous):

That's what I ended up getting for the second part. Thank you!

OpenStudy (anonymous):

you're welcome

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